# D^2 I /dt^2 + 4 dI/dt + 2504I = 110

1. Oct 5, 2013

### s3a

1. The problem statement, all variables and given/known data
The problem:
The electric current in a certain circuit is given by $d^2 I/dt^2$ + 4dI/dt + 2504I = 110. If I = 0 and dI/dt = 0 when t = 0, find I in terms of t.

The solution is attached in the TheProblemAndSolution.jpg file (along with the problem again).

2. Relevant equations
I(t) = c_1 exp((a + ib)t) + c_2 exp((a – ib)t) = c_3 exp(at) cos(bt) + c_4 exp(at) sin(bt)

3. The attempt at a solution
I found the general solution to the homogeneous equation (that does not involve the particular solution). My confusion is about the particular solution. I know that I must find the particular solution and add it to the general solution to the homogeneous version of this differential equation but, I'm a little confused as to how I am supposed to know that I(1) = 110/2504 is the initial value condition (so that I can use it to find the particular solution and, then, the general solution to the entire differential equation)?

Any input would be greatly appreciated!

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2. Oct 5, 2013

### Ray Vickson

Look at the DE for K(t) = I(t) -c, where c = constant.

3. Oct 5, 2013

### s3a

I feel that that should give me an epiphany but, I just don't see it. :(

4. Oct 5, 2013

### Ray Vickson

So, what happens when you try it?

5. Oct 5, 2013

### s3a

I get d^2 I /dt^2 + 4 dI/dt + 2504I – 110 = 0 … are you trying to get me to see that this is a homogeneous equation? If it is a homogeneous equation then, why is there a particular solution?

6. Oct 5, 2013

### LCKurtz

The solution says nothing about $I(1)$. You are looking for a particular solution of the form $I_p = C$, a constant. When you plug $I=C$ into your NH equation, you get $0 + 0 +2504C = 110$. Solve for $C$.

7. Oct 5, 2013

### Ray Vickson

You need to find the DE for K. Otherwise, why bother?

8. Oct 5, 2013

### s3a

Sorry, I double-posted.

9. Oct 5, 2013

### s3a

Thanks for the responses but, I DO get that; my problem is that if I plug in I = 0 into the differential equation because the problem claims that I = 0 (and not 110/2504) when t = 0, I would get 0 + 0 + 0 = 110, which is not correct. I do see that I = 110/2504 for t > 0, though.

My issue is what happens at t = 0 compared to t > 0.

I hope my confusion is more clear now.

Edit:
Could it be that when t = 0, I must use the homogeneous equation instead of the non-homogeneous one? If so, why would that be the case?

10. Oct 5, 2013

### LCKurtz

You have the general solution$$I(t) = e^{-2t}(A\cos(50t)+ B\sin(50t)) + \frac {110}{2504}$$
You apply the initial conditions to that equation to solve for $A$ and $B$ to get the solution to the IVP.

11. Oct 6, 2013

### s3a

Okay, I think I finally see it!

The whole I = 0 and dI/dt = 0 when t = 0 is for the general solution (only), I(t), which you/LCKurtz just gave me in your/his latest post such that the following holds, right?:

I(0) = exp(-2*0) (-110/2504 cos(50*0) + 11/6260 sin(50*0)) + 110/2504
I(0) = -110/2504 + 110/2504 = 0

P.S.
A = -110/2504
B = -11/6260

12. Oct 6, 2013

### Ray Vickson

Since you did not use the "quote" button, it is impossible to tell which message you are replying to. However, since you now have the solution, I might as well spell it out: suppose you have a general linear DE of the form
$$a_n(x) y^{(n)}(x)+ a_{n-1}(x) y^{(n-1)}(x) + \cdots + a_0 y(x) = k,$$
where $a_0$ and k are constants (with $a_0 \neq 0$). If you set $y(x) = c + z(x)$ with c=constant, the DE for z becomes
$$a_n(x) z^{(n)}(x)+ a_{n-1}(x) z^{(n-1)}(x) + \cdots + a_0 z(x) + a_0 c = k,$$
so becomes homogeneous if you choose $c = k/a_0$. That is a bog-standard trick that is used everywhere.

13. Oct 6, 2013

### s3a

Thanks for showing me that general form ... I was replying to the post right above mine (which is what I meant by “latest post”).

Furthermore, I know this is pedantic of me but, you said it was a “trick” but, it's okay to see it as a particular solution added to the general solution of the homogeneous equation instead of seeing it as a “trick” that turns a non-homogeneous equation into a homogeneous equation, right? In other words, both are correct methods of interpretation, right?

14. Oct 6, 2013

### Ray Vickson

Yes, it is equivalent as long as the last term a_0*y(x) has a constant coefficient a_0.

15. Oct 6, 2013

### s3a

Alright, thank you very much LCKurtz and Ray Vickson!

16. Oct 6, 2013

### HallsofIvy

Staff Emeritus
I presume you know that the solutions to a "homogeneous differential equation with constant (real) coefficients" must be of the form $Ce^{at}$ (where "a" is a real root of the characteristic equation), $e^{at}(Acos(bt)+ Bsin(bt))$ (where "a+ bi" is a complex root of the characteristic equation and "a- bi" is another), $t^n e^{at}$ (where "a" is a multiple real root of the characteristic equation), or $t^n e^{at}(cos(bt)+ sin(bt)$, (where "a+ bi" is a multiple root of the characteristic equation).

It can be shown (and usually is in an introductory differential equations class) that if the "non-homogenous" part is itself one those, the "particular solution" is also of that form, with "undetermined coefficients". Here, the "non-homogeneous" part is a constant so of the form "$Ce^{0t}= C$ so you try "I(t)= C" as Ray Vickson suggested, you do NOT get "d^2 I /dt^2 + 4 dI/dt + 2504I – 110 = 0" as you responded. Since I is a constant, "d^2 I/dt^2" and "4 dI/dt" are both 0 so you get only 2504I- 110= 0 and, solving for I, I= 100/2504= 50/1252= 25/626.

Another method that works for "non-homogeneous" terms of any form is "variation of parameters" (also usually taught in an "introductory differential equations course"). That method is considerably more complicated which is why so many examples in an introductory course are of type that "undetermined coefficients can be used for!

17. Oct 6, 2013

### s3a

Thanks for the supplementary information, HallsofIvy. :)

18. Oct 6, 2013

### LCKurtz

Exactly. I felt that that was what was confusing you.